Calculate: $\lim\limits_{x\to 2}\frac{\sqrt{x-1}-1}{x^2-4}$ After substituting $x$ with $y^2$ I obtain
$$\frac{y^2-2}{(y^2-2)^2}$$
and I'm not sure that this is the correct way of cancelling the square root,so can anyone help me with this problem please?
 A: If you put $y^2$ in place of $x$ you get $\sqrt{y^2-1}-1$ in the numerator.  This certainly is not $y^2-2$.
If you let $y=\sqrt{x-1}$ then you have $y^2=x-1$ and hence $x=y^2+1$.  As $x$ approaches $2$, $y$ approaches $1$.  So
$$
\lim_{x\to2} \frac{\sqrt{x-1}-1}{x^2-4} = \lim_{y\to 1}\frac{y-1}{(y^2+1)^2-4}.
$$
Since this becomes $0/0$ as $y$ approaches $1$, you need to write
$$
\lim_{y\to 1}\frac{y-1}{(y-1)(\cdots\cdots)} = \lim_{y\to 1}\frac 1 {(\cdots\cdots)}.
$$
Figure out how to complete that factorization and then go on from there.
A: $$\lim_{x \to 2}\frac{\sqrt{x-1}-1}{(x-2)(x+2)}=\lim_{x \to 2}\frac{\sqrt{x-1}-1}{(x-2)(x+2)}\frac{\sqrt{x-1}+1}{\sqrt{x-1}+1}=\lim_{x \to 2}\frac{x-2}{(x-2)(x+2)
(\sqrt{x-1}+1)}$$
$$=\lim_{x \to 2}\frac{1}{(x+2)(\sqrt{x-1}+1)}=\frac{1}{8}$$
A: Well if you substitute $y^2=x$ then you should be getting $\lim_{y\to \sqrt 2} \frac{\sqrt{y^2-1}-1}{y^4-4}$ which is quite far from what you wrote.
What you might want to consider is multiplying top and bottom by conjugate of the top.
A: Friend, I am under the impression you are unaware of L'Hopital's Rule. 
If you have a indeterminate form only of the forms $\frac{ \infty }{ \infty }$ and $\frac{ 0 }{ 0}$ then you differentiate the numerator and denomitator in order to create a function that is able to be evaluated using limits, this is considered the value of the limit. 
$$\frac{d}{dx}  \sqrt{x-1} -1= \frac{1}{2} (x-1)^{ \frac{-1}{2} } $$
And, 
$$\frac{d}{dx} (x^2+4)=2x$$
Then we must evaluate the following limit,
$$\lim_{x \rightarrow 2}  \frac{ (x-1)^{-1/2} }{4x}$$
So simple evaluation will tell us, 
$$\lim_{x \rightarrow 2}  \frac{ (x-1)^{-1/2} }{4x} =  \pm  \frac{1}{8}$$ 
Hope I helped you evaluate this limit. You must be a undergraduate in Calculus I with little knowledge of L'Hopital's Rule and other secrets to evaluate complicated limits.
